Question

measurements show that the energy of a mixture of gaseous reactants increases by 384. kj during a certain chemical reaction, which is carried out at a constant pressure. furthermore, by carefully monitoring the volume change it is determined that - 157. kj of work is done on the mixture during the reaction. calculate the change in enthalpy of the gas mixture during the reaction. round your answer to 3 significant digits. is the reaction exothermic or endothermic? exothermic endothermic

Explanation:

Step1: Recall first law of thermodynamics

$\Delta U = q + w$
Here, $\Delta U = 384\ \text{kJ}$ (energy increase of system), $w = -157\ \text{kJ}$ (work done on system is negative if work is done by system; wait correction: work done on the mixture is positive for system, wait no—standard sign: work done on system is +w, work done by system is -w. The problem says "-157 kJ of work is done on the mixture"—so work done on system is $w = -157\ \text{kJ}$? No: wait, first law: $\Delta U = q + w$, where $w$ is work done on the system. If the problem states "-157 kJ of work is done on the mixture (system)", then $w = -157\ \text{kJ}$.

Step2: Calculate heat (q) at constant pressure

At constant pressure, $q = \Delta H$. Rearrange first law: $q = \Delta U - w$
$q = 384\ \text{kJ} - (-157\ \text{kJ}) = 384 + 157 = 541\ \text{kJ}$
So $\Delta H = 541\ \text{kJ}$

Step3: Determine reaction type

Positive $\Delta H$ means endothermic.

Answer:

541 kJ
endothermic